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Pegasystems Exam PEGAPCSA87V1 Topic 8 Question 58 Discussion

Actual exam question for Pegasystems's PEGAPCSA87V1 exam
Question #: 58
Topic #: 8
[All PEGAPCSA87V1 Questions]

You are configuring duplicate case search logic in a case type. How do you ensure that resolved cases are not evaluated as potential duplicates?

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Suggested Answer: A, B

by Royal at Jan 16, 2025, 12:12 AM

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Lettie
15 days ago
Option C all the way! Gotta love that weighted condition - it's like a little 'resolved' forcefield to keep those pesky closed cases at bay.
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Phillip
1 months ago
Option C is the way to go. Checking the work status on both ends and giving 'Resolved' a weight of 0 is a smart way to avoid false positives.
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Jolene
1 days ago
Definitely, it's important to consider all possible scenarios when configuring duplicate case search logic.
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Karan
8 days ago
Definitely, it's important to carefully configure the duplicate case search logic to avoid any issues with resolved cases.
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Minna
8 days ago
That makes sense, by giving 'Resolved' a weight of 0, it ensures that it won't match as a potential duplicate.
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Myra
13 days ago
That makes sense, by giving 'Resolved' a weight of 0, it ensures that it won't match as a potential duplicate.
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Raylene
16 days ago
I agree, option C seems like the best way to make sure resolved cases are not considered duplicates.
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Muriel
22 days ago
I agree, option C seems like the best way to make sure resolved cases are not considered duplicates.
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Earleen
1 months ago
I'd go with Option C as well. It's the only one that addresses the work status of both the current and potential duplicate cases.
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Noah
1 months ago
Option C seems like the most comprehensive approach. Evaluating the work status of both the current and existing cases, and weighting 'Resolved' as 0, ensures that resolved cases are not considered duplicates.
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Fernanda
2 months ago
Hmm, that makes sense too. I guess it depends on how the logic is implemented in the case type.
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Frederick
2 months ago
I disagree, I believe the correct answer is D. We should check the potential duplicates, not the current case.
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Fernanda
2 months ago
I think the answer is A, because we need to make sure the current case is not resolved.
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